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In the first step, it involves constructing a sequence of open sets $\{U_n \}$ indexed by some re-sequencing of rational numbers $\{P_i \}$ of a topological space $X$ with the property, when $p<q$, that :

$$ U_p \subset U_q$$

(1)For constructing these sets, we arrange the set of all rational numbers in the set $\left[0,1 \right]$ into an infinite sequence in with the first two entries being $1$ and $0$. Munkres' say this is done for convenience, but I don't understand how this is so. To clarify more, my doubt is why Munkre's went out of his way to make $1$ and $0$ the first two terms.

Now , another issue , (2)later $P_i$ is considered as consisting of the first $n$ rational number in the sequence. For the set $P_{n+1} = P_n \cup \{r \}$, it is said that $r$ has an immediate predecessor $p$ in $P_{n+1}$ and an immediate succesor $q$ in $P_{n+1}$. This defies my common sense, because, what if $r$ is larger than every other element in $P_n$ or small than? (Related)

Also, my last doubt is regarding how this proof helps us actually prove the proof of the statement of Urysohn lemma.

Urysohn's lemma: Let $X$ be a normal space, let $A$ and $B$ be disjoint closed subsets of $X$ . Let $\left[a,b \right]$ be a closed interval in the real line. Then there exists a continous map

$$ f: X \to \left[ a, b \right]$$

Such that $f(x)=a$ for every $ x \in A$ , and $f(x) = b $ for every $ x \in B$

Now here is the definition of function given in step-3:

$$ f(x) = \text{inf} Q(x) = \text{inf} \{ p | x \in U_p \}$$

(3)Now, the possible set of values for $p$ is actually rational number. So, how does the infimum help turn the function into a real valued one? Maybe this is trivial to understand if one had intuition for inf but I think some more details may help.

Btw I have linked every helpful MSE post on the Urysohn Lemma's proof based on Munkres onto this MSE post

tryst with freedom
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  • Do you understand that the set of rational numbers is countable? And that every subset of a countable set is also countable? If you do, then question 1 should have an immediate answer. – Lee Mosher Aug 23 '22 at 14:50
  • My concern is mainly with $1$ and $0$ being first two entries. It being a sequence of rational number is fine. @LeeMosher – tryst with freedom Aug 23 '22 at 14:52
  • Well, in that case, start with any sequence that enumerates the rational numbers in $[0,1]$. And then, like any two terms of any sequence, you can reorder the sequence to put $1$ and $0$ as the first two terms. – Lee Mosher Aug 23 '22 at 18:16
  • I mean my question is why is it convenient to put 1 and 0 as first two terms? @LeeMosher – tryst with freedom Aug 23 '22 at 21:25
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    Then you should perhaps clarify the wording. You can see from my comments and from the one answer (so far) that your wording on 1 is difficult to understand. – Lee Mosher Aug 23 '22 at 21:59
  • This post might be helpful. – user264745 Aug 29 '22 at 08:08

2 Answers2

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For (1): the rationals are countably infinite, as are those within $[0,1]$. Formally that means there is some given bijection $\psi:\Bbb N\to\Bbb Q\cap[0,1]$ (in fact, there are loads!). We don’t really care about the exact nature of $\psi$, but Munkres wants an easy notational access to $0,1$. As $\psi$ bijects, there are distinct and unique naturals $n,m$ with $\psi(n)=0,\psi(m)=1$. There are also, as $\psi$ injects, distinct rationals $p=\psi(1),q=\psi(2)$. Let $\varphi:\Bbb N\to\Bbb Q\cap[0,1]$ be given as follows: $$k\overset{\varphi}{\mapsto}\begin{cases}0&k=1\\1&k=2\\p&k=n\\q&k=m\\\psi(k)&\text{otherwise}\end{cases}$$You can see that $\varphi$ is a bijection. So, we can enumerate the rationals in $[0,1]$ by the sequence $\{r_n\}_{n\in\Bbb N}$, with law $r_n:=\varphi(n)$, and the first two entries of this sequence are $0,1$. In this manner, you can arrange the elements of a countable enumeration to appear in a lot of different, prescribed orders. I could, arbitrarily, demand the $123456789$th element of the sequence to be $1/(123456789)$, just for fun.

For (2): the question does not make sense to me. If $r$ is the $(n+1)$th element of the sequence, then $r$ has a predecessor in $P_n$, namely the $n$th element of the sequence, but it doesn’t have a successor in $P_{n+1}$ since the next, $(n+2)$th element would be in $P_{n+2}$... perhaps I misunderstand what you mean by successor/predecessor. Please elaborate on how Munkres uses this idea, so I can see what is actually meant.

For (3): two points. Firstly, any rational-valued function can automatically be viewed as real valued. Secondly, this function is not actually rational-valued. If I understand the context correctly, it is a continuous real function - long ago, I asked about Royden’s construction of the same function, and I will link that post if you wish. The infimum of some set of rationals might be irrational: consider $\inf_{r\in\Bbb Q}\{r>\sqrt{2}\}=\sqrt{2}$. What this function is doing (perhaps Munkres differs notably from Royden on this point, I don’t know without further details from you - but I doubt it), is getting a handle on the smallest open set $U_p$ which contains the element $x$. However, such a “smallest” $U_p$ might not exist among the collection of $U_p$, as is often the case when one deals with (continuous) infinities (what is the smallest rational $x$, $x>1$? No such $x$ exists). So, we take a limit, an infimum, and it may happen (and does happen) that the value is sometime irrational. The $U_p$ are arranged such that this limiting process produces a nice continuous function with image $[0,1]$. As a concrete example, consider the context of the Euclidean subspace $[0,1]$. If the $U_p:=[0,p)$ for rational $p\in[0,1)$, and $U_1:=[0,1]$, then what is: $$\inf_p\{1/e\in U_p\}$$It is in fact just $1/e$. Indeed, the function $f(x)$ thus constructed would end up being the identity map.

FShrike
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    Thanks. I'll tidy up 2 when I get on my pc again. – tryst with freedom Aug 23 '22 at 15:18
  • " then r has a predecessor in Pn, namely the nth element of the sequence, but"... hmm they didn't say the sequence is kept strictly increasing afaik – tryst with freedom Aug 23 '22 at 15:44
  • Also in step-3 how do we know that every irrational has such a set for it? Also in step-1, my question is regarding those two were kept as first two elements not on the numeration. I understand that you can numerate rationals with naturals well – tryst with freedom Aug 23 '22 at 15:46
  • @TrystwithFreedom I suspect “predecessor of $r$” as meaning: “the previous element to $r$ in the enumeration” (i.e., if $r$ is element $n$, we ask for element $(n-1)$) not “largest element of the enumeration, smaller than $r$”. Because, for any rational $r$, there is no such thing as a $q\in\Bbb Q$, with $q<r$ and $p\in\Bbb Q,p<r\implies p\le q$. But again, please explain how Munkres uses the notion of predecessor and successor, because I am having to guess. With regards to step 1, you ask how Munkres can have $0,1$ as the first elements. I explicitly gave you a way to arrange this. – FShrike Aug 23 '22 at 15:59
  • @TrystwithFreedom For your comment regarding $3$. We know that $f$ is continuous, and that it attains $0$ and $1$. By the intermediate value theorem, it necessarily attains every element $0<t<1$, whether or not $t$ is irrational. However, there is no “corresponding set” for an irrational - that is precisely why an infimum must be taken, to make this work. The $U_p$ are rational-indexed, so no irrational $t$ can find itself a $U_t$. However, it can find a sequence of $U_{p_1},U_{p_2},...$ where $p_k\to t$ from above, which might help you make sense of it. – FShrike Aug 23 '22 at 16:03
  • No... my question is not how he put 0,1 as first .I appreciate you giving the way to construct the sequence, but my question was why is he doing it? How is it convenient – tryst with freedom Aug 24 '22 at 11:55
  • I think I need to think more of step -3. And I will check the step-2 thing. I guess it may be because he is using the word in different sense as you said @FShirke – tryst with freedom Aug 24 '22 at 11:56
  • @TrystwithFreedom Ah I see. Well, I can’t speak for Munkres without seeing their original text (I don’t have a copy) but I can tell you that Royden, Royden created the $U_p$ by induction. The induction had to begin with $U_0,U_1$ for convenience: it’s convenient because it’s just... nice. The human mind appreciates beginning at zero and ending at one. If we began at $1/40$, it might be hard to visualise where $U_0$ ends up. The $U_0,U_1$ played the important role of covering the two disjoint closed sets that define the Urysohn function. – FShrike Aug 24 '22 at 12:06
  • @TrystwithFreedom In fact, there’s nothing special whatsoever about using $[0,1]$. We could just as well use $[-123456789,1/\lfloor e^e^e^e\rfloor]$. It’s just that $[0,1]$ is nice – FShrike Aug 24 '22 at 12:09
  • @FShrike predecessor and successor notion used to describe order in $P=\Bbb{Q}\cap [0,1]$. $r=x_i$, for some $i\geq 3$. So $\exists p,q\in P_n$ such that $0\leq p\lt r\lt q\leq 1$. See this post, it is not a proof by induction. – user264745 Aug 29 '22 at 08:20
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    @user264745 I never claimed it was a proof by induction. I just wanted Tryst to explain how the terms were being used, in order for their question to be answerable. Imho Munkres makes it too complicated: Paul Frost’s answer needed a Zorn-type argument. In Royden, we make do with dyadic rationals, which has an easy induction argument, and we are done very quickly. – FShrike Aug 29 '22 at 13:36
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The convenience in (1) is that it ensures that you needn't worry about (2). I.e. $r$ cannot be larger, or smaller, than all the elements of $P_n$ because they include 0 and 1. Without that assumption he would need to include the cases where $r$ is extremal - it's easily done but unnecessary.

David Hartley
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